# It is posible for x to be equal to sinx ?

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To solve this problem we first calculate sin x / x for lim x--> 0,

This can be done by writing sin x in the Taylor series form

sin x = x - x^3/ 3! + x^5/ 5!...

So sin x / x = 1- x^2/ 3! + x^4/ 4!...

Now for lim x ---> 0 all the terms with x become 0 and we are left with 1.

So sin x/ x = 1 for the limit x---> 0.

Therefore for lim x-->0, as sin x / x = 1, it implies sin x = x.

Yes, it is possible for x = 0. But let's prove it!

We'll have to solve the equation x - sin x = 0.

The equation is a transcedental one, so we'll have to differentiate the function f.

Before differentiating, we'll check if the function is continuous. Because f(x) is formed by elementary functions as the linear one, x , and the trigonometric function, sin x, f(x) is a continuous function.

We'll differentiate f(x).

df/dx = 1-cosx

We notice that f(x) is a monotone increasing function.

( -1<cosx<1), so the difference 1 -cos x>0 =>f(x)>0, so f(x) is an injection.

We can also do a very simple calculus:

f(0)=0-sin0=0-0=0.

Because f(x) is an one-to-one, x=0 is the only solution for the equation x-sinx=0.

**So, yes, x = sin x, for x = 0.**

No, sinx < less than x for allx.

But sinx/x = 0 asx --> 0.

Suppose we draw a circle with ce ntre as origin O and rardius r. . Le t P be any point and LeX'OX be the x axis cutting the circle at X' on left and X on right.

Draw a perpenicular from P to A.

Let x = angle XOP.

AP =OP sinx < arch XP = OP x.

Divide by OP both sides:

Therefore sinx < x

Then by definiton